2. Summing up to TT (50 pts.) Write a parallel program pie.c in Cor C++ (pie.cc)

Question

2. Summing up to TT (50 pts.) Write a parallel program pie.c in Cor C++ (pie.cc) for Linux that computes an approximation of the number Tu using a series' with N+1 terms. The series sum is partitioned in T non-overlapping partial sums, each computed by T separate child processes created with the fork0library function. This program demonstrates data parallelism and interprocess communication using pipes. Each child process could perform a (potentially long computation on a separate CPU (or core), thus, the entire computation could reach a speedup close to T. Numbers N and Tare passed from the shell to the pie program as command line parameters. The main(argc, argvll) function call gets these parameters T, N) in argvlil and argv12] as strings, respectively. (Element argvloj contains the program name and we don't need it.) The first parameter, N, is the upper limit for i in the formula above. The second parameter,Tis the number of child processes that compute partial sums. N should always be greater than N T otherwise the program should display an error message and call exit(1) Use the Nilakantha approximation formula for Tu (htt n wiki wiki/Pi ia 3 4/2*3*4) -4/(4*5*6) 4/(6*7*8) 4/(8*9* 10) k*4/((2*i)* (2*i+1)*( 2*i+20)+. where k 1 if i is odd and k 1 if i is even and i goes from 1 to N. The program can be run like this from the shell: /pie NT For instance, ./pie 100 4 This command computes the approximation with 101 terms (starting with term 3) with 4 child processes. Here is a description of the algorithm to be implemented. The parent process iterates with an index variable of type int called j from 0 toT(0sj*). During iteration with index j, the parent pr 1. creates a pipe associated with child process with index j 2. creates a child process with fork0 3. writes the values N, T and j to the pipe

Explanation / Answer

int main(int argc, char *argv[]) {

if (argc != 1) {

printf("Program only takes one argument - nth tern in the Nilakantha series to use");

return -1;

}

int precision = (int) argv[0]

if (precision <= 0) {

printf("%s", "Non-negative input required");

return -1;

}

double sum = 3;

int i = 1;

for (i <= precision; i++) {

sum += nthNilakanthaNum(i);

}

}

double nthNilakanthaNum(int n) {

double denominator = 2 * n;

int i

for (i = 0; i < 2; i++) {

denominator *= denominator + 1;

}

double value = 4 / denominator;

if ((n & 1) ^ 1) { //n is odd

return value * -1;

}

return value;

}

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